Let $R$ be the region in the first and second quadrants that is inside the polar curve $r=4$ and inside the polar curve $r=4-2\sin(2\theta)$, as shown in the graph. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{2} \int_{\scriptsize\dfrac{\pi}{2}}^{\pi}16\,d\theta+\dfrac{1}{2} \int_{0}^{\scriptsize\dfrac{\pi}{2}}\left(4-2\sin(2\theta)\right)^2\,d\theta$ (Choice B) B $\dfrac12 \int_{0}^{2\pi}\left(16-\left(4-2\sin(2\theta)\right)^2\right)d\theta$ (Choice C) C $\dfrac12 \int_{0}^{\pi}\left(16-\left(4-2\sin(2\theta)\right)^2\right)d\theta$ (Choice D) D $\dfrac{1}{2} \int_{\pi}^{2\pi}16\,d\theta+\dfrac{1}{2} \int_{0}^{\pi}\left(4-2\sin(2\theta)\right)^2\,d\theta$
Solution: Since we are dealing with two separate polar curves, a good first step is to identify two areas, each enclosed by a single curve, that together define $R$. Such are $R_1$ and $R_2$ : $y$ $x$ $ R_1$ $ R_2$ $ 1$ $ 1$ $R_1$ is enclosed by $r=4$ and $R_2$ is enclosed by $r=4-2\sin(2\theta)$. Once we express them as integrals, we can find $R$ using the following relationship: $\text{Area of }R=\text{Area of }R_1+\text{Area of }R_2$ $R_1$ is enclosed by $r=4$ between $\alpha=\dfrac{\pi}{2}$ and $\beta=\pi$ : $\begin{aligned} \text{Area of }R_1&=\dfrac{1}{2} \int_{\scriptsize\dfrac{\pi}{2}}^{\pi}\left(4\right)^2\,d\theta \\\\ &=\dfrac{1}{2} \int_{\scriptsize\dfrac{\pi}{2}}^{\pi}16\,d\theta \end{aligned}$ $R_2$ is enclosed by $r=4-2\sin(2\theta)$ between $\alpha=0$, and $\beta=\dfrac{\pi}{2}$ : $\begin{aligned} \text{Area of }R_2&=\dfrac{1}{2} \int_{0}^{\scriptsize\dfrac{\pi}{2}}\left(4-2\sin(2\theta)\right)^2\,d\theta \end{aligned}$ Now we can express the area of $R$ : $\begin{aligned} &\phantom{=}\text{Area of }R \\\\ &=\text{Area of }R_1+\text{Area of }R_2 \\\\ &=\dfrac{1}{2} \int_{\scriptsize\dfrac{\pi}{2}}^{\pi}16\,d\theta+\dfrac{1}{2} \int_{0}^{\scriptsize\dfrac{\pi}{2}}\left(4-2\sin(2\theta)\right)^2\,d\theta \end{aligned}$